3. Let $b+c-a=2 x, c+a-b=2 y, a+b-c=2 z$, then $x, y, z \in \mathbf{R}^{+}$, and $a=y+z, b=x+z, c=x+y$, so the original inequality is equivalent to $\frac{x-z}{2 x}+\frac{y-x}{2 y}+\frac{z-y}{2 z} \leqslant$ 0, which means $\frac{z}{x}+\frac{x}{y}+\frac{y}{z} \geqslant 3$, obviously true.